in the given figure, ABCD is a square. if angle PQR=90 and PB=QC=DR, prove that QB=RC, PQ=QR and angle QPR=45

 

(a)

ABCD is a square

BC = CD

Given QC = RD

∴ BC - QC = CD - RD

⇒ BQ = CR

(b)

In ∆PBQ and ∆QCR

PB = QC  [given]

BQ = CR  [Proved in (a)]

∠PBQ = ∠QCR  [Each is 90°]

∴ ∆PBQ and ∆QCR [SAS Congruency]

⇒ PQ = QR, ∠BPQ = ∠CQR, ∠BQP = ∠CRQ  [C.P.C.T.]

(c)

BQC is a straight line

∴ ∠BQP + ∠PQR + ∠CQR = 180°

⇒∠BQP + ∠PQR + ∠BPQ = 180°

⇒(∠BQP + ∠BPQ) + ∠PQR = 180°

⇒180° - ∠PBQ + ∠PQR = 180° [Angle sum property for ∆PBQ]

⇒180° - 90° + ∠PQR = 180°

⇒∠PQR = 90°

(d)

∆PQR is an isosceles right angled triangle

∴ ∠QPR = 45°

Cheers!


 

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